Help with Vector problem

Two vectors a and b have the components, in meters, ax = 3.5, ay = 1.1, bx = 0.9, by = 4.2.
(a) Find the angle between the directions of a and b.
60°

There are two vectors in the xy plane that are perpendicular to a and have a magnitude of 4.5 m. One, vector c has a positive x component and the other, vector d, a negative x component.

(b) What is the x component of c?
m
(c) What is the y component of c?
m

(d) What is the x component of d?
m
(e) What is the y component of d?
m

- I have tried numerous things to figure out the xy components on this problem, but I don't think I really even know where to start. If somebody could point me in the right direction and give me some hints on how to solve this, that would be great. This is the only problem of the 14 on my homework assignment that I can't seem to figure out. I'm trying to get this submitted by 2am CST. Thanks for any help.

Two vectors a and b have the components, in meters, ax = 3.5, ay = 1.1, bx = 0.9, by = 4.2.
(a) Find the angle between the directions of a and b.
60°

There are two vectors in the xy plane that are perpendicular to a and have a magnitude of 4.5 m. One, vector c has a positive x component and the other, vector d, a negative x component.

(b) What is the x component of c?
m
(c) What is the y component of c?
m

(d) What is the x component of d?
m
(e) What is the y component of d?
m

- I have tried numerous things to figure out the xy components on this problem, but I don't think I really even know where to start. If somebody could point me in the right direction and give me some hints on how to solve this, that would be great. This is the only problem of the 14 on my homework assignment that I can't seem to figure out. I'm trying to get this submitted by 2am CST. Thanks for any help.

If you know the "dot product" then remember that c dot a must be zero for them to be perpendicular.

Another related hint is that given the slope of a vector is the y component over the x component, the slope of vector c will be -1 over the slope of vector a given they are perpendicular (only works in 2-dim where dot product above works in any number of dimensions)

Third hint: You'll note they say two vectors c and d satisfy the given properties. Visualize vector a and ask why two distinct vectors perpendicular to it? How are these two related? If you weren't in 2 dimensions i.e. the plane how would this change the problem?

as well as how to determine when 2 vectors are perpendicular to each other?

- I've tried solving for x and then plugging back into either aXc or a.c. I've also tried plugging back into the formula for finding the magnitude and a couple of other things that I don't even think are mathematically correct.

I know how to get the angle between a and b from the first part of the problem by using a.b=|a||b|cost. I used the dot product of to get the magnitude of a and b so that I could find the angle between them (60).

I'm not completely sure how to determine if 2 vectors are perpendicular or not, but I know from the information given to the second part of the problem that c and d are perpendicular to a. Also, I know that d would be the opposite of c, so I only need to find out what cx and cy are, but I'm really not sure of how to get to that point.

Looking through my scribble I've done trying to solve it. I had already determined that a.c was = zero. I was mostly trying to use substitution to figure out x and y. I used cx and cy for my variables and came up with 3.5cx + 1.1cy=0 and then solved for cx to get 3.18cy=cx. I tried plugging that back into a few other equations, but couldn't ever get any kind of an answer. Is that on the right track at all? Also, didn't get the impression that vector b had much to do with the second portion of the problem (since it only mentions a in relation to c and d).

ok back to what jambaugh said draw the vector a, and draw c that's perpendicular to it. see anything interesting? are the components of a sort of the components of c?
and also draw another vector oposite of c, how could a's components be related to d?

Also if you're given a vector c/d but you need it in a specific length but same direction as c/d then you solve for a new vector v = m(c or d/length of c or d) where m is the given length that they want.

ok back to what jambaugh said draw the vector a, and draw c that's perpendicular to it. see anything interesting? are the components of a sort of the components of c?
and also draw another vector oposite of c, how could a's components be related to d?

Also if you're given a vector c/d but you need it in a specific length but same direction as c/d then you solve for a new vector v = m(c or d/length of c or d) where m is the given length that they want.

- To be honest, I'm not sure if it's from a huge lack of sleep or not, but I'm just not grasping this one for some reason. I gather that the components of c would more than likely be somewhat close to those of a since they are perpendicular, but I'm not quite sure the relation of them. I'm not sure why I'm struggling with this problem so much.

just draw the vector, draw another 1 that's sort of perpendicular to it, and you'll see something about it, that it's components are opposite of a's but one of the components is negative. because there will be 2 perpendicular vectors to any vector in the x-y plane.

just draw the vector, draw another 1 that's sort of perpendicular to it, and you'll see something about it, that it's components are opposite of a's but one of the components is negative. because there will be 2 perpendicular vectors to any vector in the x-y plane.

OK, Good. If you add the z direction so you have 3 dimensions you'll get a whole plane of vectors perpendicular to a. Of those only two are in the x-y plane.

You should start this problem by writing out using variable for the components the vector for which you are searching: [itex] \mathbf{c} = c_x \mathbf{x}+ c_y\mathbf{y} = (c_x,c_y)[/itex]

You've got that "orthogonal" means zero dot product with a so formally take the dot product and you have a linear equation. (with two unknowns).
But you also have magnitude information which gives you another quadratic equation. You have two equations and two unknowns. Solve and you're done.

To solve use the linear equation to express one of the variables in terms of the other and substitute this into the quadratic. You should get a very simple quadratic equation (with no degree-1 term) so that you get the +/- solutions.

If you work out some numbers but aren't sure if you made an error (or even if you are) test them again with the conditions specified in the problem. Verify the magnitude and the orthogonality condition.

I don't consider that I've solved a problem like this until I'm willing to bet cash money that I have the correct answer. Being sure you're right is part of having it "solved".

- Thanks for all the help, but I never did figure out the problem before it was due. I talked with my professor and felt pretty dumb with how easy the problem actually is.

To solve the second part (finding xy components), I just had to find the angle of a with respect to the x-axis (tan^-1(1.1/3.5) which gave me 17.45 degrees. Then since c and d are perpendicular to a, I could find their angles with respect to the x-axis. For c it was 72.55 degrees and so I took 4.5cos72.55 for the x component and then 4.5sin72.55 for the y component. I just took the opposites of those for d. So it was way easier than I was making it out to be.

Two vectors a and b have the components, in meters, ax = 3.5, ay = 1.1, bx = 0.9, by = 4.2.
(a) Find the angle between the directions of a and b.
60°

There are two vectors in the xy plane that are perpendicular to a and have a magnitude of 4.5 m. One, vector c has a positive x component and the other, vector d, a negative x component.

(b) What is the x component of c?
m
(c) What is the y component of c?
m

(d) What is the x component of d?
m
(e) What is the y component of d?
m